Total number of possible events = 2^5 = 32
Frequency of exactly 3 heads (HHHT*, THHHT, *THHH) = 2+1+2 = 5
Frequency of exactly four consecutive heads (HHHHT, THHHH) = 2
Frequency of five consecutive heads = 1
Frequency of required events = 5+2+1 = 8
Required probability = 8/32 = 1/4
1&2) You are working on a prediction problem. You need to solve $$\pi(k\ge450|4\text{ success of 10 tries})=1516927277253024\sum_{k=450}^{1000}\int_0^1\binom{1000}{k}p^k(1-p)^{1000-k}(1-p)^{25}p^{23}\mathrm{d}p$$
That is your posterior probability that the next 1000 tosses will result in 450 or more successes. You should gamble no more than the prize times the probability.
3) No, choosing the prior that you believe to be closest to the truth is the most scientific solution. You should use any real information that you have. You can test the sensitivity of your result to your prior, but since you are gambling what you need to do is work out the knowledge that you really have.
If you believe it is a nearly fair coin then $B(20,20)$ as a prior is quite reasonable, though $B(2,2)$ is as well.
EDIT
You are solving $$\pi'(k=K|X)=\int_0^1f(k=K|p)\pi(p|X)\mathrm{d}p$$ for each value of $k\ge{450}$, where $f$ is the likelihood function, $\pi$ is the posterior and $\pi'$ is the prediction. You are removing the uncertainty regarding $p$ by marginalizing it out. You have to sum all the cases in your hypotheses space which is $k\ge{450}$.
You are looking at the likelihood of every outcome weighted by the posterior probability over the entire set of all possible parameter values.
Best Answer
Here the natural null-hypothesis $H_0$ is that the coin is unbiased, that is, that the probability $p$ of a head is equal to $1/2$. The most reasonable alternate hypothesis $H_1$ is that $p\ne 1/2$, though one could make a case for the one-sided alternate hypothesis $p>1/2$.
We need to choose the significance level of the test. That's up to you. Two traditional numbers are $5$% and $1$%.
Suppose that the null hypothesis holds. Then the number of heads has *binomial distribution with mean $(900)(1/2)=450$, and standard deviation $\sqrt{(900)(1/2)(1/2)}=15$.
The probability that in tossing a fair coin the number of heads differs from $450$ by $40$ or more (in either direction) is, by symmetry, $$2\sum_{k=490}^{900} \binom{900}{k}\left(\frac{1}{2}\right)^{900}.$$ This is not practical to compute by hand, but Wolfram Alpha gives an answer of roughly $0.008419$.
Thus, if the coin was unbiased, then a number of heads that differs from $450$ by $40$ or more would be pretty unlikely. It would have probability less than $1$%. so at the $1$% significance level, we reject the null hypothesis.
We can also use the normal approximation to the binomial to estimate the probability that the number of heads is $\ge 490$ or $\le 410$ under the null hypothesis $p=1/2$. Our normal has mean $450$ and variance $15$ is $\ge 490$ with probability the probability that a standard normal is $\ge 40/15$. From tables for the normal, this is about $0.0039$. Double to take the left tail into account. We get about $0.0078$, fairly close to the value given by Wolfram Alpha, and under $1$\%. So if we use $1$\% as our level of significance, again we reject the null hypothesis $H_0$.
Comments: $1$. In the normal approximation to the binomial, we get a better approximation to the probability that the binomial is $\ge 490$ by calculating the probability that the normal is $\ge 489.5$. If you want to look it up, this is the continuity correction. If we use the normal approximation with continuity correction, we find that the probability of $490$ or more or $410$ or fewer heads is about $0.008468$, quite close to the "exact" answer provided by Wolfram Alpha. Thus we can find a very accurate estimate by, as in the bad old days, using tables of the standard normal and doing the arithmetic "by hand."
$2$. Suppose that we use the somewhat less natural alternate hypothesis $p>1/2$. If $p=1/2$, the probability of $490$ or more is about $0.00421$. Thus again at the $1$% significance level, we would reject the null hypothesis, indeed we would reject it even if we were using significance level $0.005$.
Setting a significance level is always necessary, for it is possible for a fair coin to yield say $550$ or more heads in $900$ tosses, just ridiculously unlikely.